Find $\dfrac{∂f}{∂x}$ and $\dfrac{∂f}{∂y}$ for$$f(x, y)=\int_{x+y}^{x-y}\sin t^3\,\mathrm dt.$$
I looked around before posting this and got a solution but I am not sure if it is right. I got that $\dfrac{∂f}{∂x}$ and $\dfrac{∂f}{∂y}$ were both $3\sin(x-y)(x-y)^2 - 3\sin(x+y)(x+y)^2$ using this post as a guide.
Is that correct?
No, that's not correct. When we apply the chain rule here, the only things we're differentiating are the limits of the integral. The derivative never goes far enough inside to hit that $t^3$. In fact, to remove the temptation there? Let's just call the function we're integrating $g(t)$, and plug in the formula for it at the last possible moment. $$\frac{\partial}{\partial x}\int_{x+y}^{x-y}g(t)\,dt = g(x-y)\cdot\frac{\partial}{\partial x}(x-y) - g(x+y)\cdot \frac{\partial}{\partial x}(x+y) = g(x-y)-g(x+y)$$ And then, $g(x-y)-g(x+y)=\sin\left((x-y)^3\right)-\sin\left((x+y)^3\right)$. That's the partial derivative in $x$. I leave the other as an exercise, to show that you understand this.